Question
One mole of an ideal gas undergoes a process$$p=\frac{p_{0}}{1+\left(V / V_{0}\right)^{2}}$$where $p_{0}$ and $V_{0}$ are constants. Find the temperature of the gas when $V=V_{0}$.
Step 1
Step 1: We are given the pressure equation as $$ p=\frac{p_{0}}{1+\left(V / V_{0}\right)^{2}} $$ We can multiply both sides by $V$ to get $$ pV=\frac{p_{0}V}{1+\left(V / V_{0}\right)^{2}} $$ Show more…
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One mole of an ideal gas undergoes a process: $$ P=\frac{P_{o}}{1+\left(V_{0} / V\right)^{2}} $$ Here $P_{a}$ and $V_{n}$ are constants. Change in temperature of the gas when volume is changed from $V=V_{n}$ to $V=2 V$ is (a) $-\frac{2 P_{n} V_{o}}{5 R}$ (b) $\frac{11 P_{e} V_{a}}{10 R}$ (c) $-\frac{5 P_{e} V_{o}}{4 R}$ (d) $P_{o} V_{e}$
One mole of an ideal gas with heat capacity at constant pressure $C_{P}$ undergoes the process $T=T_{0}+\alpha V$, where $T_{0}$ and $\alpha$ are constants. If its volume increases from $V_{1}$ to $V_{2}$, the amount of heat transferred to the gas is (A) $C_{P} R T_{0} \ln \left(\frac{V_{2}}{V_{1}}\right)$ (B) $\alpha C_{P} \frac{\left(V_{2}-V_{1}\right)}{R T_{0}} \ln \left(\frac{V_{2}}{V_{1}}\right)$ (C) $\alpha C_{P}\left(V_{2}-V_{1}\right)+R T_{0} \ln \left(\frac{V_{2}}{V_{1}}\right)$ (D) $R T_{0} \ln \left(\frac{V_{2}}{V_{1}}\right)+\alpha C_{P}\left(V_{1}-V_{2}\right)$
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The Derivative
Product and Quotient Rules
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